of 1 56
The Solar System As A Solution To The Wave Equation!
By!
Ian Beardsley!
Copyright © 2024"
of 2 56
Contents
Abstract…………………………………………………………………….3
1.0 The Lunar Formulation………………………………………………4
2.0 The Solar Formulation………………………………………………9
3.0 Equating The Lunar And Solar Formulations
Yield Our 1 Second Base Unit………………………………………….13
4.0 Explicitly Deriving Our Equations And Formulating Spin……..17
5.0 Formulating Angular Momentum………………………………24
6.0 Why Might The Historical Second Be
A Physical Constant?……………………………………………………..35
7.0 Solutions For Jupiter And Saturn………………………………38
8.0 Modeling Habitable Star Systems………………………………42
9.0 A Genesis Project……………………………………………………..47
10.0 Hydrocarbons The Backbones of Life Chemistry…………….51!
Appendix 1…………………………………………………………………….53
Data Used In This Paper…………………………………………………55
of 3 56
Abstract
We find we there exists a solution to the wave equation for the Solar System, the wave equation
that solves the atom, that centers around the Earth, the third planet, where life is abundant. It
exists in terms of the Moon and a base unit of 1 second. The Moon seems to be some kind of a
Natural yardstick and the old mystery of the Moon perfectly eclipsing the Sun as seen from the
Earth because of its size and distance from the Earth, plays a key part in the solution. Though
one second developed historically from ancient times from the Babylonians who got it from the
Sumerian base 60 sexagesimal counting, it turns out to be a base unit that solves the planets and
atoms (the proton) in terms of one another. This has archaeological implications, because it
might take us to the realm of theories of Ancient Aliens. In this study we see the Earth/Moon/
Sun system is an elegant, dynamic structure, that is complexly functional, which leaves us
wondering what kind of forces could be behind its origins in that random chance seems
improbable.
The Moon, of the Earth, has long been thought of as not making sense, it is large and massive for
a moon of a terrestrial planet while it has a very low density, low mass for its size, when it should
be comparable to that of the Earth. When we dropped the launch pods from lunar craft for the
Apollo mission on the Moon and measured the seismic activity, it seemed to ring like a bell for
an hour, something characteristic of something hollow. Further studies were done by NASA
after that in a classified study called Chapel Bell, which remains classified 50 years after we went
to the Moon. It has been suggested by some scientists that the Moon would make more sense if
it was hollow, and even that it was a hollow spacecraft put there for the purpose of making life
more successful on Earth. If it was a craft, and we could enter it and move it to adjust our
climate when it is going awry not only from global warming but due to periodic ice ages caused
by small cyclical changes in the Earth orbit, that would solve a lot.
Included in this study is modeling not just the Earth/Moon/Sun system but habitable star
systems in general in terms of them having an enigmatic moon like we have. While it may not be
mainstream science to suggest the Moon of the the Earth is hollow, it is a part of it to suggest
that very successful habitable planets would need to have an enigmatic moon like we do. I even
find that our description of the atom’s proton in terms of a base unit of one second from which
we derive the Earth’s moon, that the equation is based around hydrocarbons, the skeletons of
life chemistry.
of 4 56
1.0 The Lunar Formulation
I find I can find a solution for the orbit of the Earth around Sun with the Schrödinger wave
equation that has the solution to an electron around a proton in the atom if I quantize it
according to the Earths’ Moon, and a base unit of one second. The second was developed in
ancient times not in terms of the atom or things like the mass of the Sun or the Earth, but I find
it does happen to be the right unit of time for our solution, and I will remark later on why that
might be. The Schrödinger wave equation is, in spherical coordinates
1.1
Its solution for the atom is
1.2
1.3
is the energy for an electron orbiting protons and , is the orbital shell for an electron with
protons, the orbital number. I find the solution for the Earth around the Sun utilizes the
Moon around the Earth. This is different than with the atom because planets and moons are not
all the same size and mass like electrons and protons are, and they don’t jump from orbit to
orbit like they do. I find that for the Earth around the Sun
1.4
1.5
is the kinetic energy of the Earth, and is the planet’s orbit. is the radius of the Sun,
is the radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is
the orbit number of the Earth which is 3 and is the Planck constant for the solar system,
which we will presently formulate. Instead of having protons, we have the radius of the
Sun normalized by the radius of the Moon. It is has always been an amazing fact the the sizes of
the Moon and the Sun are such that given their orbital distances, the Moon as seen from the
Earth perfectly eclipses the Sun. This becomes part of the theory and we suggest it is a condition
for sophisticated planets that harbor life by writing it:
1.6
That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is
about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say
the system is quantized by the Moon and the base unit of one second.
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sin θ
∂
∂θ
(
sin θ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
r
n
=
n
2
ℏ
2
Z k
e
e
2
m
e
E
Z
r
n
Z
n
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
K E
e
r
n
R
⊙
r
m
M
e
M
m
n
ℏ
⊙
Z
R
⊙
/R
m
r
planet
r
moon
≈
R
star
R
moon
of 5 56
1.7
Then wavelength associated with the Moon divided by the speed of light should be 1 second if
our planetary system is quantized in terms of the Moon and one second. We have
1.8
And we see it is, so we have
1.9
Now we derive the solar Planck constant, which is a Planck constant for the the planetary
system. We do it with one second because equation 1.9 suggest we should. We suggest it is such
that it is given by the standard Planck constant for the atom, , times some constant, , and the
Kinetic energy of the Earth.
1.10
1.11
Where
1.12
And, this works. We find it in fact derives from the atom. I had found that in terms of the radius
of a proton and its mass, I could derive one second as equations 1.11 and 1.12 already do:
1.13
Here is how we arrive at the planetary Planck Constant from the radius of a proton. Energy is
given by Planck’s constant and frequency
We have
We take the rest energy of the mass of a proton :
λ
moon
=
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22kg)
3
= 3.0281E8m
λ
moon
c
=
3.0281E8m
299,792,458m /s
= 1.010secon d s
λ
moon
c
= 1secon d
h
C
ℏ
⊙
= (hC )K E
e
hC = 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996354secon d s
E = h f
f = 1/s, h = J ⋅ s, h f = (J ⋅ s)(1/s) = J
E = J = Joules = en erg y
m
p
of 6 56
The frequency of a proton is
This gives
The radius of a proton is then
Equation 1.14
This is close to the CODATA value for the proton radius around 2018 (0.842E-15m). Let us now
suggest there is some Planck-type constant for the Earth-Sun system that is connected to the
Planck constant for quanta we will call it and denote as . We will say is connected to
by some constant and is given by the kinetic energy of the Earth . That is, we have
If
We find
satisfies this condition.
We write
Or,
1.15
We multiply equation 14 by to get
E = m
p
c
2
f
p
=
m
p
c
2
h
m
p
c
2
h
⋅
r
p
c
=
2
3
≈ ϕ =
m
p
c
h
r
p
m
p
r
p
=
2
3
⋅
h
c
r
p
=
2
3
⋅
h
cm
p
h
h
⊙
h
h
p
h
⊙
h
p
C
K E
e
h
⊙
= (h
p
C )KE
e
h
p
C = 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
h
p
C = t
1
t
1
=
1
3
⋅
h
p
α
2
c
1
3
⋅
2π r
p
G m
3
p
r
p
/r
p
r
2
p
=
2
3
⋅
hr
p
cm
p
of 7 56
=
Multiply the radicand by to get
Multiply the radicand by to get
=
Multiply outside the radical by to get
This factors into
1.16
We have
1.17
Because of equation 1.15:
, , , , QED.
r
p
=
2
3
⋅
hr
p
cm
p
2
hr
p
cm
p
⋅
1
6
h
h
⋅
m
2
p
m
2
p
⋅
c
c
r
p
= 2
h
2
m
2
p
c
2
⋅
r
p
c
m
3
p
h
⋅
1
6
2π G /2πG
r
p
= 2
h
2
m
2
p
c
2
⋅
r
p
c
m
3
p
h
⋅
2π G
12π G
2h
m
p
c
r
p
c
m
3
p
h
⋅
2π
3
⋅
1
4π
⋅
G
G
α
2
/α
2
r
p
=
2h
α
2
c
⋅ α
2
m
p
1
3
⋅
2π r
p
G m
3
p
⋅
Gc
4πh
r
p
=
1
3
⋅
h
α
2
c
1
3
⋅
2π r
p
G m
3
p
(
18
3
α
2
m
p
Gc
4πh
)
r
p
= 6α
2
m
p
Gc
4πh
⋅ t
1
t
1
=
1
3
⋅
h
p
α
2
c
1
3
⋅
2π r
p
G m
3
p
t
1
= 1secon d
h
p
C = t
1
ℏ
⊙
= (hC )K E
e
of 8 56
Now we derive the value of our solar Planck constant
=
=
=
=
One of the things that becomes important later is that:
1.18.
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
1
3
⋅
18769
299792458
1
3
⋅
2π (0.833E − 15)
(6.67408E − 11)(1.67262E − 27)
3
1.55976565E 33
s
m
m
kg
3
⋅
s
2
kg
m
3
=
s
m
s
2
kg
2
m
2
=
s
m
⋅
s
kg ⋅ m
=
1
kg
⋅
s
2
m
2
1
C
= kg
m
2
s
2
=
1
2
mv
2
= en erg y
hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon d s ≈ 1.0secon d s
hC =
(
kg
m
s
2
⋅ m ⋅ s
)
(
1
kg
⋅
s
2
m
2
)
(
kg
m
2
s
)(
1
kg
⋅
s
2
m
2
)
= secon d s
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
ℏ
⊙
= (hC )K E
earth
= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s
K E
moon
K E
earth
(Ear th Da y) = 1secon d
of 9 56
2.0 The Solar Formulation
Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the
mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the
Sun. In our lunar formulation we had:
2.0.
We remember the Moon perfectly eclipses the Sun which is to say
2.1.
Thus equation 2.0 becomes
2.2.
The kinetic energy of the Earth is
2.3
Putting this in equation 2.2 gives the mass of the Sun:
2.4
We recognize that the orbital velocity of the Moon is
2.5.
So equation 2.4 becomes
2.6
This gives the mass of the Moon is
2.7.
Putting this in equation 2.0 yields
2.8.
K E
e
= 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
r
e
r
m
K E
e
= 3
r
e
r
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
e
=
1
2
⋅
GM
⊙
M
e
r
e
M
⊙
= 3r
2
e
⋅
GM
e
r
m
⋅
M
3
m
ℏ
2
⊙
v
2
m
=
GM
e
r
m
M
⊙
= 3r
2
e
⋅ v
2
m
⋅
M
3
m
ℏ
2
⊙
M
3
m
=
M
⊙
ℏ
2
⊙
3r
2
e
v
2
m
K E
e
=
R
⊙
R
m
⋅
G
2
M
2
e
M
⊙
2r
2
e
v
2
m
of 10 56
We now multiply through by and we have
2.9
Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular
momentum quantized as Bohr discovered for the atom by The angular
momentum we will call , the subscript for Planck. We have
We write for the solution of the Earth/Sun system:
2.10
Let us compare this to that of an atom:
2.11
We notice that in equation 2.10
; ; ; ;
is really . We can write 2.10 as
2.12
We say. . That is
Let us see how accurate our equation is:
M
2
e
/M
2
e
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2r
2
e
v
2
m
M
2
e
ℏ
⊙
h /2,2h /2,3h /2,...
L
p
p
L
p
= r
e
v
m
M
e
= r
e
v
m
M
e
= (1.496E11m)(1022m /s)(5.972E 24kg) = 9.13E 38kg ⋅
m
2
s
L
2
p
= r
2
e
v
2
m
M
2
e
= 7.4483E 77J ⋅ m
2
⋅ kg = 8.3367E 77kg
2
⋅
m
4
s
2
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
E = −
Z
2
n
2
⋅
k
2
e
e
4
m
e
2ℏ
2
Z
2
n
2
→
R
⊙
R
m
k
2
e
→ G
2
e
4
→ M
4
e
m
e
→ M
⊙
ℏ
2
→ L
2
p
L
p
ℏ
⊙
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
ℏ
⊙
= ℏ
⊙
/2π
ℏ
⊙
= 9.13E 38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39J ⋅ s
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
of 11 56
=
=
We have that the kinetic energy of the Earth is
Our equation has an accuracy of
Which is very good.
We call the same in our solar solution. But we now want our solution for the solar
formulation. is the mass of the Earth and is the mass of the Sun. We have our solution
might be!
2.13 !
Where is !
We have
2.14
This has an accuracy of
!
Thus the solutions to the wave equation!
R
⊙
R
m
(6.67408E − 11)
2
(5.972E 24kg)
4
(1.9891E 30kg)
2(8.3367E 77kg
2
⋅
m
4
s
2
)
R
⊙
R
m
(6.759E 30J )
R
⊙
R
m
=
6.957E8m
1737400m
= 400.426
K E
e
= 2.70655E 33J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.70655E 33J
2.7396E 33J
= 98.79 %
L
p
ℏ
⊙
r
n
M
e
M
⊙
r
n
=
ℏ
2
⊙
GM
3
e
⋅
R
m
R
⊙
h
⊙
ℏ
⊙
= 9.13E 38J ⋅ s
r
3
=
(9.13E 38)
2
(6.67408E − 11)(5.972E 24kg)
3
⋅
1
400.5986
= 1.4638E11m
1.4638E11m
1.496E11m
100 = 97.85 %
of 12 56
!
For the solar formulations are!
2.12.
2.13.
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sinθ
∂
∂θ
(
sinθ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
r
n
=
ℏ
2
⊙
GM
3
e
⋅
R
m
R
⊙
ℏ
⊙
= 9.13E 38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39J ⋅ s
of 13 56
3.0 Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit
Let us equate equation 1.4 with equation 2.12:
1.4
2.12
This gives:
3.1
We remember that
This gives
3.2
We have
3.3
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
ℏ
⊙
= (hC )K E
p
hC = 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
r
e
v
m
M
e
=
1
6α
2
⋅
r
p
m
p
h 4π
Gc
⋅
1
2
M
e
v
2
e
⋅
M
2
e
M
⊙
M
3
m
3
2v
m
=
v
2
e
r
e
(1secon d )
M
2
e
M
⊙
M
3
m
3
of 14 56
This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in
terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,
and the orbital number of the Earth. Let us compute
3.4
Let us see how well equation 3.3 works. at aphelion is 966 m/s and .
. We have
That is an accuracy of
Equation 3.3 can be written:
3.5
From equation 1.13:
We have
3.6
Since , the diameter of the Earth orbit, we have
3.7
And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice
versa. We basically have
M
2
e
M
⊙
M
3
m
3
=
(5.972E 24kg)
2
(1.9891E 30kg)
(7.34763E 22kg)
3
(1.732)
= 321,331.459 ≈ 321,331
v
m
v
e
= 29,800m /s
r
e
= 1AU = 1.496E11m
2(966m /s) =
(29,800m /s)
2
1.496E11m
(1sec)(321,331.459)
1,907m /s = 1,932m /s
1907
1932
= 98.7 %
1secon d = 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 2r
e
v
m
v
2
e
⋅
M
3
m
3
M
2
e
M
⊙
2r
e
= d
e
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= d
e
v
m
v
2
e
⋅
M
3
m
3
M
2
e
M
⊙
of 15 56
3.8
Where is the Earth orbital number. We have
= 7.83436E4seconds
EarthDay=(24)(60)(60)=86400 seconds,
accuracy
This last equation, equation 41, we can use to find the rotation period, or the length of the day,
of an earth-like planet in the habitable zone of any star system, so it would be very useful.
We want to turn our attention to equation 3.5 and write it
3.9
We see equating solar and lunar formulations for energy yield the base unit of one second.
Equating the lunar and solar solutions for orbitals instead of the for energies, which we just did,
we have
Yields
1secon d = 2v
m
r
e
v
2
e
M
3
m
3
M
2
e
M
⊙
1secon d =
K E
moon
K E
earth
Ear th Da y
1secon d =
M
m
v
2
m
M
e
v
2
m
Ear th Da y
Ear th Da y =
2r
e
v
m
M
m
M
⊙
n
n = 3
Ear th Da y =
2(1.496E11m)
966m /s
7.34763E 22kg
1.9891E 30kg
1.732
7.834E4s
86,400s
100 = 90.675 %
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
=
L
2
p
GM
3
e
⋅
R
m
R
⊙
of 16 56
3.10
Where . The accuracy of this is
Thus the energy equations gave the equation 3.1:
3.1. !
And equating the orbital equations gives
3.11
These last two yield
3.12
The accuracy is
Is 98% accuracy."
ℏ
2
⊙
M
3
e
M
3
m
=
3
2
L
2
p
R
2
m
R
2
s
3/2 = cos(π /6)
(2.8314E 33)
2
(5.972E 24)
3
(7.34763E 22)
3
= (0.866)(9.13E 38)
2
(1737400)
2
(6.96E8)
2
4.29059E 72 = 4.5005E 72
4.29059E 72
4.5005E 72
100 = 95.3 %
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
L
2
p
=
2 3
3
⋅
M
3
e
M
3
m
R
2
⊙
R
2
m
⋅ ℏ
2
⊙
2
R
⊙
R
m
M
e
M
⊙
= 1
(400.5986)
5.972E 24kg
1.9891E 30kg
= 1
0.98 = 1
of 17 56
4.0 Explicitly Deriving Our Equations And Formulating Spin
The Bohr model of the atoms was formulated before the Schrödinger wave equation, but it is the
solution to the Schrödinger wave equation for the atom. Let’s see how Niels Bohr arrived at it,
then arrive at it for the planets. In the Bohr Model of the atom Niels Bohr made three postulates:
1. Electrons exist in stable orbits at discrete distances from the nucleus.
2. The angular momentum of the electron is quantized according integer multiples of the
reduced Planck constant.
3. Electrons only gain and lose energy by jumping from one orbit to another.
Thus to begin we equate the centripetal force of the electron to the force between two charges
given by Coulomb’s law:
4.1
This gives a velocity for the electron of
4.2
The angular momentum of the electron is quantized by integer multiples of the reduced Planck
constant:
4.3
Putting the velocity from equation 4.2 into equation 4.3:
4.4
Z=1 is the smallest radius possible for r and is the Bohr radius:
Putting the equation for v in the equation for kinetic energy:
4.5
And, into that the equation for r, we get
4.6
m
e
v
2
r
= Z
k
e
e
2
r
2
v = Z
k
e
e
2
m
e
r
m
e
vr = nℏ
r
n
=
n
2
ℏ
2
Z k
e
e
2
m
e
r
1
=
ℏ
2
Z k
e
e
2
m
e
= 5.29E − 11m = 52.9pm
E =
1
2
Mv
2
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
= −
13.6Z
2
n
2
eV
of 18 56
We now write equation 1 but not using Coulomb’s law but rather Newton’s universal law of
gravitation, We have
4.7
We see that the orbital velocity of the Moon is
4.8
The angular momentum of the Moon is
4.9
Putting the orbital velocity of the Moon from equation 4.8 into equation 4.9 we have
4.10
We know planetary systems are not like atomic systems, planets aren’t all the same size like
electrons are and like protons are, which have the same charge but planets don’t have the same
gravity because they don’t have the same mass either. Nor do planets jump from one orbit to
another like electrons do. Z is the atomic number, the number of protons at the nucleus of the
atom, but for planetary systems there is only the Sun at the center. We find that for this to work
with the that we formulated in this theory, that
And,…
Where for the Earth n=3 and further that the Moon near perfectly eclipses the Sun as seen from
the Earth meaning that
M
m
v
2
m
r
m
= G
M
e
M
m
r
2
m
v
m
=
GM
e
r
m
M
m
v
m
r
m
= nℏ
⊙
M
m
GM
e
r
m
r
m
= nℏ
⊙
M
2
m
GM
e
r
m
⋅ r
m
= n
2
ℏ
2
⊙
r
m
=
n
2
Z
⋅
ℏ
2
⊙
GM
2
m
M
e
ℏ
s
n
2
→
1
n
Z →
R
m
R
⊙
r
e
r
m
=
R
⊙
R
m
of 19 56
We see Z, the proton in the atom is the radius of the Sun, but it is divided by the radius of the
Moon to put it in Moon-units, that the Moon is like a natural yardstick. We further have that
So, our final equation is
4.11
Let us go back to equation 5 to get the equation for the energy of the Earth in its orbit and the
equation for the velocity equation 4.8…
And we get
4.12
Putting equation 4.11 for into this
4.13
We find that and we write
4.14
And we find we multiply this by and we have
4.15
The Bohr model of the atom was later reinterpreted by deBroglie where he said Bohr’s condition
that angular momentum be an integer multiple of is really that it is the condition of a standing
wave where
4.16
That electrons behave as waves and that the wavelength of the electron is
M
e
→ M
m
r
3
=
R
⊙
R
m
⋅
ℏ
2
⊙
GM
3
m
⋅
1
3
E =
1
2
Mv
2
v
m
=
GM
e
r
m
E =
1
2
M
GM
e
r
m
r
3
E =
1
2
M
GM
e
⋅ GM
3
m
ℏ
2
⊙
M → M
e
E =
G
2
M
2
e
M
3
m
2ℏ
2
⊙
n
R
⊙
R
m
E
3
= 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
ℏ
n λ = 2π r
of 20 56
4.17
Which is to say
4.18
Or, that
4.19
Where is angular momentum, the same units as which is . So we have
4.20
Which is Bohr’s second postulate that angular momentum is quantized in integer multiples of
the reduced Planck constant.
Now, with what we have done up to this point finding the wave length of the Moon, a lot begins
to unfold. We have said in this theory:
Thus with what we arrived at
And with
We have
4.21
λ =
h
mv
nh
mv
= 2π r
nh
2π
= mvr
mvr
h
Joules ⋅ secon d s
L =
nh
2π
ℏ
⊙
= (hC )K E
p
hC = 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
K E
m
K E
e
(Pl an et Day) = 1.0secon d s
λ
moon
c
= 1secon d
K E
moon
1.0secon d s
(Ear th Da y) = K E
earth
ℏ
⊙
= (1secon d )K E
earth
of 21 56
4.22
4.23
4.24
4.25
4.26
Thus the planet day, or spin, is given by
4.27
Where
4.28
This has to do with the concept of spin in quantum mechanics. The Earth day can be thought of
as spin. This is the rotation period of the Earth, which is different than angular momentum of
something orbiting something, it is more the angular momentum of something spinning on its
axis, but this is a messy subject in quantum mechanics. It was found not just the orbit of an
electron was necessary to describe, but as well the spin of an electron on its axis. However, when
you describe the rotation of an electron classically, its spin becomes faster than the speed of
light, so it doesn’t work. It became necessary to introduce spin not as something that is actually
happening, but as an abstract concept that represents something that is not concrete. So they
came up with the concept of spin numbers, numbers that can only be given by spin, s=n/2 where
n is an integer.
Similarly, spin for planets becomes difficult to apply. Planets close into the Sun, like Venus, are
tidally locked in with the Sun, their rotation (or spin) being greatly slowed down by the Sun’s
gravity. The rotation of Venus, closer in than the Earth, has a rotation (day) of 243 days. The
Earth of course is 24 hours. Mars is close to the Earth day. The Earth sidereal day is 23 hours 56
minutes 4.09 seconds, the Mars sidereal day is 24 hours 37 minutes 22 seconds. The Jupiter day
is 9 hours 56 minutes, and the Saturn day is 10 hours 34 minutes. This has to do with the spin
concept because
(K E
moon
)(Ear th Da y) = ℏ
⊙
ℏ
⊙
=
λ
moon
c
⋅ K E
earth
K E
earth
=
ℏ
⊙
1secon d
K E
m
K E
e
(Pl an et Day) =
λ
moon
c
λ
moon
=
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22kg)
3
= 3.0281E8m
(Pl an et Day)
3
=
K E
3
K E
m
(1secon d )
K E
1
= M ercur y, KE
2
= Venus, K E
3
= Ear th, KE
4
= Mars, . . .
K E
m
K E
e
(Pl an et Day) =
ℏ
2
⊙
GM
3
m
⋅
1
c
(Pl an et Day)
3
=
K E
3
K E
m
(1secon d )
of 22 56
We have to find x for x-seconds, the number of seconds that describe the planet’s day. We can’t
consider the spins of Mercury, and Venus because their orbits are so slowed down by tidal forces
from the Sun, Venus even spins in the opposite direction of its orbit. It starts working at Earth
orbit, where life is successful. Equation 27 is the case for the Earth and works quite well. We
postulate that since the gas giants, Jupiter and Saturn, carry the better part of the solar system’s
mass they embody the dynamics behind the idea behind the solar system. We start with Jupiter,
and its largest Moon, Ganymede. It is the third significant Moon around Jupiter and we
postulate the third orbit is where something significant happens, as the with Earth, the third
planet from the Sun. Ganymede has an orbital velocity of 10.880 km/s and its mass is 1.4819E23
kg. Its eccentricity is very low (0.0013) meaning like our Moon has an almost perfectly circular
orbit. We have its kinetic energy is
We have for Jupiter its day is
(9 hours)(60)(60)+56(60)=35,760 seconds
The kinetic energy of Jupiter is
We have
Spin in quantum mechanics is given by s=n/2 where n is an integer giving 1/2, 1, 3/2, 2, 5/2.
Thus spin of Earth is 1 (1 second) we skip Mars (3/2), go to Jupiter is 2 (2 seconds), so Saturn
should be 5/2 (2.5 seconds).
Titan is the largest moon of Saturn and the second largest in the solar system. It is mass is
1.3452E23 kg and its orbital velocity is 5.57 km/s. We have its kinetic energy is
K E
1
= M ercur y, KE
2
= Venus, K E
3
= Ear th, KE
4
= Mars, . . .
K E
G
=
1
2
(1.4819E 23kg)(10,880m /s)
2
= 8.77095E 30Joules
1
2
(1.898E 27kg)(13,060m /s)
2
= 1.61865E 35Joules
(Pl an et Day)
5
=
K E
5
K E
G
(x − secon d s)
35,760s =
1.61865E 35J
8.77095E 30J
(x − secon d s)
35,760s = (18,454.67)x
x = 1.9377secon d s ≈ 2secon d s
(Pl an et Day)
3
=
K E
3
K E
m
(1secon d )
(Pl an et Day)
5
=
K E
5
K E
G
(2secon d s)
of 23 56
We have for Saturn its day is
(10 hours)(60)(60)+(34)(60)=35,760 seconds
The kinetic energy of Saturn is
Exactly what we predicted. We have:
We can’t consider Venus, its orbit is so slowed down from tidal forces of the Sun. We have
Mars has no Moons that we can use for the calculation, they are small and irregular, and
perhaps captured passing asteroids. This concept of spin, worked very nicely for the planets.
K E
T
=
1
2
(1.3452E 23kg)(5,570m /s)
2
= 2.08673E 30Joules
1
2
(5.68319E 26kg)(10,180m /s)
2
= 2.945E 34Joules
35,760s =
2.945E 34J
2.08673E 30J
x
35,769s = (14,113)x
x = 2.53447secon d s ≈ 2.5secon d s
(Pl an et Day)
3
=
K E
3
K E
m
(1secon d )
(Pl an et Day)
5
=
K E
5
K E
G
(2secon d s)
(Pl an et Day)
6
=
K E
6
K E
T
(5/2secon d s)
n /2 = 2/2 = 1secon d(Ear th)
n /2 = 3/2 = Mars
n /2 = 4/2 = 2secon d s(Jupiter)
n /2 = 5/2 = 2.5secon d s(Sat ur n)
of 24 56
5.0 Formulating Angular Momentum
We would like to know what spin means for the planets as analogous to spin for particles in
quantum mechanics. We recall:
Where is the Earth orbital number. We have
= 7.83436E4seconds
EarthDay=(24)(60)(60)=86400 seconds
accuracy
We consider
This gives
5.1
In General
5.2
Where subscript “moon” is not necessarily The Moon, but a moon. We might write it
5.3
For Earth we have:
5.4
Ear th Da y =
2r
e
v
m
M
m
M
⊙
3
n = 3
Ear th Da y =
2(1.496E11m)
966m /s
7.34763E 22kg
1.9891E 30kg
1.732
7.834E4s
86,400s
100 = 90.675 %
(Ear th Da y) =
K E
earth
K E
moon
(1secon d )
K E
earth
K E
moon
(1secon d ) =
2r
e
v
m
M
m
M
⊙
3
K E
n
K E
moon
(t
n
) =
2r
n
v
moon
M
m
M
⊙
n
t
n
=
2r
n
v
moon
M
m
M
⊙
n ⋅
K E
moon
K E
n
t
3
=
2r
3
v
moon
M
m
M
⊙
3 ⋅
K E
moon
K E
3
of 25 56
At perigee, maximum lunar orbital velocity the kinetic energy is 1022 m/s, so its kinetic energy
is
So
Using Ganymede with Jupiter its largest most massive moon, we have for Jupiter:
=
Using Titan with Saturn its largest most massive moon, we have for Saturn
=
5.5
5.6
5.7
t
3
=
2(1.496E11m)
966m /s
7.347673E 22kg
1.9891E 30kg
3 ⋅
3.428E 28J
2.7396E 33J
= 0.98secon d ≈ 1secon d
K E
moon
=
1
2
(7.347673E 22kg)(1022 m /s)
2
= 3.837E 28J
t
3
=
2(1.496E11m)
1022m /s
7.347673E 22kg
1.9891E 30kg
3 ⋅
3.837E 28J
2.7396E 33J
= 1.037secon d ≈ 1secon d
r
j
= (5.2AU )(1.496E11m /AU ) = 7.7792E11m
t
5
=
2(7.7792E11m)
10,880m /s
1.4819E 23kg
1.9891E 30kg
5 ⋅
8.77095E 30Joules
1.61865E 35Joules
3.162666secon d s ≈ 3secon d s
r
s
= (10.494388AU )(1.496E11m /AU ) = 1.56996E18m
t
6
=
2(1.56996E12m)
5,570m /s
1.3452E 23kg
1.9891E 30kg
6 ⋅
2.08673E 30Joules
2.945E 34Joules
16.257secon d s ≈
65
4
secon d s ≈ 16secon d s
t
3
=
2r
3
v
moon
M
m
M
⊙
3 ⋅
K E
moon
K E
3
= 1.00secon d s
t
5
=
2r
5
v
Ganymede
M
m
M
⊙
5 ⋅
K E
Ganymede
K E
5
= 3.00secon d s
t
6
=
2r
6
v
Titan
M
m
M
⊙
6 ⋅
K E
Titan
K E
6
= 16.00secon d s
of 26 56
We notice here that the velocity of the Moon cancels with one of the velocities in the kinetic
energy, leaving momentum times one half. The one have cancels with the 2 in the equation, so
we have momentum times distance in the numerator, which get divided by the Kinetic energy of
the Earth. We can do the same with all three equations, so we have
5.8
5.9
5.10
Let’s write the one for Earth:
5.11
Where n=3, the orbital number of the Earth.
5.12
We will see we are not done with this equation, momentarily. If p for the Moon and r for the
Earth are spin, angular momentum, then this spin, or angular momentum, is given in terms of
one over the orbital number to the one fourth. We see angular momentum, or spin is given not
by the Moon’s momentum and the Moon’s orbital radius, but by the Moon’s momentum and the
Earth’s orbital radius. Since one second is given by
We can substitute any of these into the Equation. We can write
5.13
t
3
=
M
m
M
⊙
3 ⋅
p
moon
r
earth
K E
earth
= 1.00secon d s
t
5
=
M
m
M
⊙
5 ⋅
p
Ganymede
r
Jupiter
K E
Jupiter
= 3.00secon d s
t
6
=
M
m
M
⊙
6 ⋅
p
Titan
r
Satu r n
K E
Satu r n
= 16.00secon d s
M
m
M
⊙
⋅ n
1
4
⋅
p
moon
r
earth
K E
earth
= 1.00secon d s
p
moon
r
earth
= 1.00secon d s ⋅
M
⊙
M
m
K E
earth
⋅ n
−
1
4
K E
m
K E
p
(Pl an et Day) = 1.0secon d s
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996354secon d s
1
3
⋅
h
α
2
c
1
3
⋅
2π r
p
G m
3
p
= 1secon d
p
moon
r
earth
= K E
moon
(Ear th Da y) ⋅
M
⊙
M
m
⋅ n
−
1
4
of 27 56
We are still not done with equation 5.13
That is an accuracy of 85.723%
Using the maximum orbital velocity of the Moon, as opposed to its average, we have
We have suggested its current orbit is optimum for life with the condition
The important thing here is the equation
Because we see angular momentum for the planets - which are not like electrons around a
proton, where electrons are all the same size and mass, and the planets are not - is formulated
not by momentum of the Moon times the Moon’s orbital radius, but by the momentum of the
Moon times the Earth orbital radius, which it travels as well, but as epicycles of the Moon
around the Sun. This is how quantization of the planets is different than with electrons around a
proton. Nor do planets jump from orbit to orbit like electrons do. We are not through with this
equation, the Earth is inclined at its rotations to the ecliptic, the path of its orbit around the Sun,
by 23.44 degrees. The Moon is inclined to the ecliptic by 5.44. We want to consider how much of
the rotational angular momentum of the Earth is in the direction of the orbital angular
momentum of the Moon. We have
K E
moon
=
1
2
(7.347673E 22kg)(1022 m /s)
2
= 3.837E 28J
(7.347673E 22kg)(1022 m /s)(1.496E11m) = (3.837E 28J )(86400s)
1.9891E 30kg
7.347673E 22kg
3
−
1
4
1.1234E 37kg
m
2
s
= (3.315E 33)(5203)(0.7598)kg
m
2
s
1.1234E 37kg
m
2
s
= 1.3105E 37kg
m
2
s
(7.347673E 22kg)(1082 m /s)(1.496E11m) = (3.837E 28J )(86400s)
1.9891E 30kg
7.347673E 22kg
3
−
1
4
1.18935E 37E 37kg
m
2
s
= 1.3105E 37kg
m
2
s
r
earth
r
moon
≈
R
Sun
R
moon
p
moon
r
earth
= K E
moon
(Ear th Da y) ⋅
M
⊙
M
m
⋅ n
−
1
4
of 28 56
5.13
Where
Now we are done with equation 5.13 It has an accuracy of 98%.
Particles have an angular momentum associated with them, but they cannot generate this
angular momenta. For instance if an electron were generating its angular momentum, it would
have to be rotating at its surface faster than the speed of light. Let’s see this. The angular
momentum of a particle is
Where s is its spin number. For an electron this is 1/2. We have
In classical mechanics the angular momentum of this electron at its equator is is given by
Thus the electron would have to be spinning at its surface with a velocity of
p
moon
r
earth
= K E
moon
(Ear th Da y) ⋅
M
⊙
M
m
⋅ n
−
1
4
cos(θ )
θ = 23.44
∘
+ 5.14
∘
= 28.58
∘
J = ℏ s(s + 1)
J =
1
2π
6.62607E − 34J ⋅ s 3/2 = 9.1326E − 35J ⋅ s
J =
2
5
m
e
r
e
v
e
=
2
5
(9.1326E − 31kg)(2.818E − 15m) = (1.0294E − 45)v
e
(1.0294E − 45)v
e
= 9.1326E − 35J ⋅ s
of 29 56
That would be 296 times the speed of light. An electron cannot be spinning this fast so we give
the electron a spin number which is 1/2, it is 1 for a photon. Then we develop a theory for the
electron that accounts for it in terms of this abstract notion of spin being a ratio of integers.
A photon has a spin of 1 so
Since we are considering the spin of the Earth in our theory, let’s compute its angular momenta:
=
4.14
Let us compute the orbital angular momentum of the Moon:
5.15
And, compare the orbital angular momentum of the Moon to the rotational angular momentum
of the Earth:
5.16
The orbital angular momentum of the Moon is almost exactly the integer 4 times greater than
the rotational angular momentum of the Earth. It is the orbital angular momentum of the Moon
that holds the Earth with its angular momentum at its tilt to the Sun making life on Earth
optimally successful by allowing for the seasons preventing extreme hot and extreme cold.
So as we said, the angular momentum of particles is not real, but represented by integers and
their ratio by way of the way it is formulated in quantum mechanics. We would have that the
angular momentum operator that acts on wave functions is!
!
Which is hermitian so it is an observable, you can measure it and you can transpose it and
complex conjugate it and you get the same matrix. We have!
!
v
e
= 8.87E10m /s =
8.87E10
299,792,458m /s
= 296c
J = ℏ s(s + 1) =
1
2π
(6.62607E − 34) 2 = 1.4914E − 34J ⋅ s
L = Iω =
2
5
M
e
R
2
e
=
2
5
M
e
R
e
v
e
2
5
(5.972E 24)(6.3781E6m)
2
= 9.7176765E 37kg ⋅ m
2
ω =
2π
86400s
= 7.27E − 5
ra d
s
L
earth
= (9.7176765E 37kg ⋅ m
2
)(7.27E − 5ra d /s) = 7.064751E 33J ⋅ s
J
moon
= mvr = (7.34767309E 22kg)(1022 m /s)(3.84E8m) = 2.88358E 34J
J
moon
L
earth
=
2.88358E 34J ⋅ s
7.064751E 33J ⋅ s
= 4.081 ≈ 4
L
z
=
xp
y
−
yp
x
[
L
x,
L
y
] = iℏ
L
z
of 30 56
Is the angular momentum commutator. But in quantum mechanics we have spin, so:!
!
We want that!
!
That is for an electron!
!
Because the spin is 1/2. The 1, 0 matrix is the state or eigenvalue, and the hbar over 2 is the
eigenvalue or measurable, observable and the square matrix is the hermetic operator is a
measurable, observable as well. Thus spin operator is:!
!
This is spin around the z axis in the x-y plane. We say a unitary operator acts on a state and
produces a state .!
!
If it acts on but rotates through an angle of zero, that is doesn’t rotate at all, then!
!
Which means is the identity matrix , Now rotate it through a very tiny angle by the very
small amount of some matrix , then we have:!
!
!
We form R dagger, that is we transpose and complex conjugate to get :!
!
and are unitary matrices so !
!
That is the condition of being a unitary matrix.!
[
S
x,
S
y
] = iℏ
S
z
S
z
[1⟩ =
ℏ
2
[1⟩
ℏ
2
(
1 0
0 −1
)(
1
0
)
=
ℏ
2
(
1
0
)
S
z
=
ℏ
2
(
1 0
0 −1
)
a
a′
R[a⟩ → [a′ ⟩
a
R
a
R[a⟩ → [a⟩
R
I
ϵ
M
R[a⟩ → (
I + iϵ
M )[a⟩
R = (
I + iϵ
M )
R
R
†
R
†
→ (
I − iϵ
M
†
)
R
R
†
R
R
†
=
I
of 31 56
!
is so small that the term with it vanishes. We have!
!
!
This is the condition for a hermitian matrix, a measurable, observable.!
!
Because it is a unitary matrix. The condition of a special unitary matrix is that its determinant is
1 and that its trace is 0. That is!
!
!
These special unitary matrices are rotation matrices, they are!
, , !
They have to have determinants 1 or -1 and traces 0. The first has determinant -1, the second
determinant +1, and the third -1. The traces of all three are zero, The special unitary matrix is
not just the operator for electron spin, but rotates up quarks to down quarks and vice versa so
it is a 2 by 2 matrix called su(2). A unitary matrix that can rotate three items, like Red, Green,
and Blue colors of quarks is a 3 by 3 matrix. Thus correspondingly these matrices are the weak
and strong forces, respectively, because rotation of these particle types is responsible for the
the weak and strong forces, where rotation is a change in state of a quark like up or down or
red and green. There is a third matrix, but it is not a unitary matrix because it is for the electric
force which uses a photon as the exchange particle, and the photon has no up or down state,
nor does it change color, it just has one state we call u(1). Thus the formulation of the standard
model of particles physics is!
su(3) X su(2) X u(1)!
We can say that the spin of a planets is given in seconds (its rotation period) instead of with
angular momentum as is done with particles in quantum mechanics. However in quantum
mechanics, this angular momentum is not due to any kind of rotation of particles we know of,
because a particle, like an electron, would have to be rotating at its surface faster than the speed
of light. So spin is described by an integer divided by 2, like for an electron is n/2=1/2, These
numbers that we associate with spin are an abstract idea that don’t pertain to anything real that
we know of, however, in quantum field theory the angular momentum is thought to be the
rotation of the field associated with particles like electrons, but this theory has not achieved
what it strives to do, which in part is reconcile relativity with quantum mechanics. In my theory
(
I + iϵ
M )(
I − iϵ
M
†
) =
I + iϵ(M − M
†
) − i
2
ϵ
2
M
M
†
=
I
ϵ
2
iϵ(
M −
M
†
) = 0
M =
M
†
R
R
†
= 1
(
a b
c d
)
= ad − bc = 1
a + d = 0
(
0 1
1 0
)
(
0 −i
i 0
)
(
1 0
0 −1
)
of 32 56
for the spin of planets, it is given in seconds, and it is a curiosity that the base unit comes out to 1
second.
5.17
5.18
5.19
We can’t consider Venus, its orbit is so slowed down from tidal forces of the Sun. We have the
following for spin
5.20
5.21
5.22
5.23
Mars has no Moons that we can use for the calculation, they are small and irregular, and
perhaps captured passing asteroids. This concept of spin, worked very nicely for the planets.
5.24
5.26
5.27
Which our spin for the planets defined as we have done we want to write the quantum
mechanical condition eigenstates, and eigenvalues as given by the Hamiltonian operator, ,
which is
5.28
5.29 We might guess we have
5.30
(Pl an et Day)
3
=
K E
3
K E
m
(1secon d )
(Pl an et Day)
5
=
K E
5
K E
G
(2secon d s)
(Pl an et Day)
6
=
K E
6
K E
T
(5/2secon d s)
n /2 = 2/2 = 1secon d(Ear th)
n /2 = 3/2 = Mars
n /2 = 4/2 = 2secon d s(Jupiter)
n /2 = 5/2 = 2.5secon d s(Sat ur n)
λ
moon
c
=
K E
moon
K E
earth
(Ear th Da y)
λ
moon
c
=
ℏ
2
s
GM
3
moon
⋅
1
c
λ
moon
c
= 1secon d
H
H [a⟩ = λ[a⟩
H =
λ
moon
c
0
0 −
λ
moon
c
of 33 56
5.31
5.32
5.33
The matrices we chose for the systems states are
Are the matrices for horizontal and vertical polarizations respectively for a photon. In the sense
that
Meaning if you prepare vertically polarized light (y) and pass it through a horizontal polarizer
(x), you have a probability of zero of it it getting through. And in the sense that
Meaning if you prepare horizontally polarized light and send it through a horizontal polarizer
you have 100% chance of it getting through. The same with vertical polarized light passing
through a vertical polarizer.If we have 45 degree polarization, then we need the matrix for
forward-slash backward-slash polarization, which we find is
The states for forward-slash 45 degree polarization and backward-slash 45 degree polarizations
are respectively:
,
[a⟩ = [x⟩, [y⟩ =
(
1
0
)
,
(
0
1
)
λ
moon
c
=
K E
moon
K E
earth
(Ear th Da y)
λ
moon
c
0
0 −
λ
moon
c
(
1
0
)
=
K E
moon
K E
earth
(Ear th Da y)
(
1
0
)
λ
moon
c
0
0 −
λ
moon
c
(
0
1
)
=
K E
moon
K E
earth
(Ear th Da y)
(
0
1
)
[a⟩ = [x⟩, [y⟩ =
(
1
0
)
,
(
0
1
)
⟨x][y⟩
2
=
(
1 0
)
(
0
1
)
= 0
2
= 0
⟨x][x⟩
2
=
(
1 0
)
(
1
0
)
= 1
2
= 1
H
⊗
=
(
0 1
1 0
)
(
1
1
)
(
1
−1
)
of 34 56
And, they are normalized
,
They have to be normalized because we must have:
So that
They are normalized so that
So that is or . Either a positive or negative outcome for the experiment, a yes or no. Thus
if vertical light is sent through a forward-slash 45 degree polarizer we have
Which is right, half the light will be polarized.
1/ 2
1/ 2
1/ 2
−1/ 2
⟨ for ward ][ for ward ⟩
2
= 1
[ for war d ⟩ =
[x⟩ + [y⟩
2
[ba ck war d ⟩ =
[x⟩ − [y⟩
2
(
0 1
1 0
)
1/ 2
1/ 2
= 1
1/ 2
1/ 2
(
0 1
1 0
)
1/ 2
−1/ 2
= − 1
1/ 2
1/ − 2
λ
1
−1
⟨ for ward ][ver t ical⟩
2
=
(
1/ 2 1/ 2
)
(
0
1
)
=
1
2
of 35 56
6.0 Why Might The Historical Second Be A Physical Constant?
We want to turn our attention to equation 3.9:
3.9
Because it may tell us why 1 second is the basic unit of our quantum mechanical solution for the
Earth/Moon/Sun system, which happens to coincide with the the fundamental unit of time we
developed since ancient times in our calendar we have today.
Let us talk, first, about how we got the unit of a second to measure time. The Sumerians, who
were of the first to settle down from wandering and gathering, from following the herds and
hunting, who fired clay to make the first homes, developed a system of writing and mathematics,
and agriculture, some 12 thousand years ago, gave us our system of measuring time.
They made their system of counting base 60, called sexagesimal, or hexagesimal, where we use
today base 10. They chose base 60 because it is evenly divisible by so much:
1,2,3,4,5,6, 10, 12, 15, 20, 30,…
and, because there are 12 months in a year, approximately the number of new Moons in a year,
and there are 5 fingers on each hand, and 12 times 5 is sixty. Twelve as well is an important
number and they wanted to factor base twelve into their counting (duodecimal) where twelve is
very abundant too, divisible evenly by
1,2,3,4,6,…
So they divided the day into 12 hours and the night into 12 hours, so the time from sunrise to
sunrise and sunset to sunset was 24 hours. So the Earth’s rotation period to this day is divided
into 24 hours. Because they had a base sixty counting system, they divided each hour into 60
minutes, and each minute into 60 seconds. If we divide the Earth’s rotation period into 24
hours, and each hour into 60 minutes, and each minute into sixty seconds, we get the duration
of a second we have today.
Looking at equation 3.9 the velocity of the Moon is given by
6.1
And the velocity of the Earth is given by
6.2
For the Earth to be in the habitable zone of the star it orbits, must be exactly what it is and
therefore must be exactly what it is. If it is a condition for life to optimize that the Moon must
perfectly eclipse the Sun as seen from the Earth it has to be that:
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
v
m
=
GM
e
r
m
v
e
=
GM
⊙
r
e
r
e
v
e
of 36 56
6.3
And, then must be exactly what it is, and therefore must be exactly what it is. Therefore the
orbital period of the Moon must be
6.4
And, the orbital period of the Earth must be
6.5
Thus the orbital period of the Moon has to be
6.6
And, the orbital period of the Earth has to be
6.7
Also the Sumerians divided the circle into 360 parts giving us the 360 degrees we have in a circle
today because 6 times 60 is 360. Thus since there are 365.25 days in a year, the time for the
Earth to revolve once around the Sun, it moves through about 1 degree per day. The Moon
moves through about . Thus we must have the Earth and lunar
orbits give one second because
Where . Because we have said that in this equation and must be exactly what they
are, which is such that the Moon moves through 12 times more degrees per day than the Earth
and there are then about 12 moons in a year because the the Earth moves through about 1
degree in a day.
The interesting thing is that we have the duration of a second is what it is not because we were
attempting to find some base unit of time for Nature, but rather we have the duration of a
second from ancient times that came to us over many years of dividing the rotation of the Earth
into hours, minutes, and seconds in such a way that we ended up with it. However we may have
had some sort of a thought going into it that was six-fold in Nature that lead it to being natural if
we consider:
6.8
And that thought may be was in a six-fold pattern that the motions of the Earth/Moon/Sun
system had in it approximately which we inadvertently characterized in the calendar:
r
e
r
m
≈
R
⊙
R
m
r
m
v
m
T
m
=
2π r
m
v
m
T
e
=
2π r
e
v
e
T
m
= 27.32d a ys
T
e
= 365.25d a ys
360deg /30d ays = 12d e g /d a y
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
d
e
= 2r
e
v
m
v
e
1secon d =
1
365.25
⋅
1
24
⋅
1
60
⋅
1
60
= 0.0000000316881year
of 37 56
6.9
Which can be written
6.10
Where there are 360 degrees in a circle and as such the Earth approximately moves through one
degree per day around the Sun. We achieved this because the ancients of Sumeria and Babylonia
from which the ancient Greeks received their counting system used base 60, sexagesimal
because 60 is evenly divisible by so much, that is by
1,2,3,4,5,6, 10, 12, 15, 20, 30,…
We have one second in the following
,
(
360da ys
year
)(
24h ours
d a y
)(
60min
h our
)(
60sec
mi n
)
= 31104000secon d s /year
12
2
(
60da ys
year
)(
1h our
d a y
)(
60min
h our
)(
60sec
mi n
)
K E
moon
K E
earth
(Ear th Da y) = 1secon d
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
hC =
1
3
⋅
h
α
2
c
1
3
⋅
2π r
p
G m
3
p
= 1.0secon d s
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
= 1secon d
λ
moon
c
=
ℏ
⊙
GM
3
moon
⋅
1
c
= 1secon d
of 38 56
7.0 Solutions For Jupiter And Saturn
We have the solution for the Bohr hydrogen atom is
7.1
is the atomic number, the number of protons orbited, but for our system it is 1, because one
body is orbited. is the orbit number, which we will deal with later. We have our solution for the
Earth/Moon/Sun system is
7.2
Let is the earth mass, is the lunar mass, and is our
Planck constant for the Earth/Moon/Sun system. We compute equation 2:
=
The kinetic energy of the Earth is
This is
The Moon perfectly eclipses the Sun which means as seen from the Earth the Moon appears to
be the same size as the Sun. This is because
7.3
Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,
and is the Moon’s radius. This is because though the Moon is 400 times further from the Sun
than it is from the Earth, the Sun is 400 times larger than the Moon. That is
Thus
7.4
We find the quantum mechanical solution to the Earth/Moon/System is
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
Z
n
E =
G
2
M
2
m
3
2h
2
⊙
n
2
M = M
e
m = M
m
h
⊙
= 2.8314E 33J ⋅ s
E =
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
n
2
3.93E 30Joules
n
2
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
1/n
2
= 697
r
e
r
m
≈
R
⊙
R
m
r
e
r
m
R
⊙
R
m
6.96E8m
1737400m
= 400.5986
R
⊙
R
m
= 400
of 39 56
7.5 !
for Earth orbit (Third Planet).!
!
!
We see the solar Planck constant we developed works in solving the Schrödinger wave equation
for the Earth/Moon/Sun system. My belief as to why this was never done is that it had to be
realized that the solution of the Earth kinetic energy around the Sun as a quantum mechanical
state is based on the Moon around the Earth.
Now we want to find what the wave equation solutions are for Jupiter and Saturn because they
significantly carry the majority of the mass of the solar system and thus should embody most
clearly the dynamics of the wave solution to the Solar System.
I find that as we cross the asteroid belt leaving behind the terrestrial planets, which are solid,
and go to the gas giants and ice giants, the atomic number is no longer squared and the square
root of the the orbital number moves from the numerator to the denominator. I believe this is
because the solar system here should be modeled in two parts, just as it is in theories of solar
system formation because there is a force other than just gravity of the Sun at work, which is the
radiation pressure of the Sun, which is what separates it into two parts, the terrestrial planets on
this side of the asteroid belt and the giants on the other side of the asteroid belt. The effect the
radiation pressure has is to blow the lighter elements out beyond the asteroid belt when the
solar system forms, which are gases such as hydrogen and helium, while the heavier elements
are too heavy to be blown out from the inside of the asteroid belt, allowing for the formation of
the terrestrial planets Venus, Earth, and Mars. The result is that our equation has the atomic
number of the heavier metals such as calcium for the Earth, while the equation for the giants has
the atomic numbers of the gasses. We write for these planets
7.6
So, for Jupiter we have (And again using the maximum orbital velocity which is at perihelion):
E = n
R
⊙
R
m
⋅
G
2
M
2
m
3
2h
2
⊙
n = 3
3(400.5986)(3.93E30J ) = 2.7268585E 33J
2.7268585E33
2.7396E33
= 99.5 %
E =
Z
n
G
2
M
2
m
3
2ℏ
2
⊙
K E
j
=
1
2
(1.89813E 27kg)(13720m /s)
2
= 1.7865E 35J
E =
Z
H
5
(6.67408E − 11)
2
(1.89813E 27kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
E =
Z
H
5
(3.971E 35J ) = Z
H
(1.776E 35J )
of 40 56
Jupiter is mostly composed of hydrogen gas, and secondly helium gas, so it is appropriate that
.
Our equation for Jupiter is
7.7
Where is the atomic number of hydrogen which is 1 proton, and for the orbital
number of Jupiter, . Now we move on to Saturn…
=
The equation for Saturn is then
7.8
It makes sense that Saturn would use Helium in the equation because Saturn is the next planet
after Jupiter and Jupiter uses hydrogen, and helium is the next element after hydrogen. As well,
just like Jupiter, Saturn is primarily composed of hydrogen and helium gas.
Our equation in this paper for the Earth orbit does not depend on the Moon’s distance from the
Earth, only its mass. The Moon slows the Earth rotation and this in turn expands the Moon’s
orbit, so it is getting larger, the Earth loses energy to the Moon. The Earth day gets longer by
0.0067 hours per million years, and the Moon’s orbit gets 3.78 cm larger per year. It is believed
the Moon came from a chunk knocked off the Earth from a collision with a Mars sized
protoplanet. At the time that the Moon formed it was about 4 Earth radii distant 4.5 billion
years ago. After 100 million years the Moon became tidally locked making its rotation period
equal to its orbital period keeping one face always towards the Earth. After 500 million years the
Moon was orbiting at about 20 Earth radii. It is believed that during the period of heavy
bombardment in a chaotic early solar system about 4.1 to 3.8 billion years ago a large object did
a close pass pulling the Moon further changing its orbit giving it its 5 degree offset from the
Earth’s equator. Our wave equation solution may only use the Moon’s mass but the equation for
Z
H
=
1.7865E 35J
1.776E 35J
= 1.006pr oton s ≈ 1.0pr oton s = hydrogen(H )
Z = Z
H
E =
Z
H
5
G
2
M
2
j
M
3
m
2ℏ
2
⊙
Z
H
n = 5
n = 5
K E
S
=
1
2
(5.683E 26kg)(10140m /s)
2
= 2.92E 34J
E =
Z
6
(6.67408E − 11)
2
(5.683E 26kg)
2
(7.347673E 22)
3
2(2.8314E 33)
2
Z
2.45
(3.5588E 34J ) = Z(1.45259E 34J )
Z(1.45259E 34J ) = (2.92E 34J )
Z = 2proton s = Heliu m(He)
E =
Z
He
6
G
2
M
2
s
M
3
m
2ℏ
2
⊙
of 41 56
kinetic energy of the Moon to kinetic energy of the Earth times the Earth Day equal to about one
second:
Which we connect with the equation where the proton gives one second
This holds for when the Moon was at a distance from the Earth such that it appears the same
size as the Sun, which means:
Which is when the two equations above for one second connect to our wave equation solution to
the Earth
Because .
The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the
Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get too
extremely hot or too extremely cold. We see the Moon may be there for a reason.
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996354secon d s
r
planet
r
moon
≈
R
star
R
moon
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
star
R
moon
=
R
⊙
R
m
of 42 56
8.0 Modeling Habitable Star Systems
We want to solve a habitable planetary system in general and apply it to another star system. We
must include the solution of its moon as well, because the Moon of the Earth makes life
optimally possible. However, our Moon is counter-intuitive, it breaks all of the rules of the other
moons in our solar system. One is it has a low mass for its size. It has been said that it could be
hollow. Even suggested that it is a hollow-spacecraft put there to make life as we know it
possible. In our quantum theory for the solar system, we may have discovered the science of the
engineers who may have made the Moon, and we are possibly applying their science here. Our
quantum theory for the solar system also has a solar solution as compared to a lunar solution, so
it works for star systems in general.
The perfect star system to use for our purposes here would be Tau Ceti. Tau Ceti is a star
spectrally similar to the Sun and the closest of such a G-class star, at about 12 light years distant.
Its mass is 78% of the Sun’s. Its mass is
Its radius is:
And its luminosity is
It has several unconfirmed planets with one potentially in the habitable zone. It has been the
subject of science fiction literature. It is in the constellation Cetus. From Tau Ceti the Earth
would be seen to be in the northern hemisphere constellation Bootes with an apparent
magnitude of 2.6. It is spectral class G8 V where our Sun is G2 V.
For the luminosity of a main sequence star we only need to know the mass of the star to get its
luminosity:
8.1
This holds in solar masses and solar luminosities. The habitable zone of a star is given by
luminosity. If the star is 100 times brighter than our Sun, then the habitable zone is 10 times
further than the Earth is from the Sun by the inverse square law.
8.2
In astronomical units and solar luminosities. From we can get the orbital period in Earth
years:
8.3
We can get the radius of the Moon of the planet if we know the mass of the planet, :
M
s
= 0.783 + / − 0.012M
⊙
R
s
= 0.793 + / − 0.004R
⊙
L
s
= 0.488 + / − 0.0010L
⊙
L = M
3.5
r
p
= L
s
r
p
T
2
p
= r
3
p
M
e
of 43 56
8.4
We want the mass of the Earth from the mass of the Sun. We would guess we can get it from the
mass of the Sun, the size of the Sun, and the base 60 sexagesimal system upon which we have
found everything is based. I find the relationship exists for the Earth:
8.5
(36)(5)=180=360/2 and 360/6=60 is base 60 sexagesimal. This gives
This is an accuracy of
We can get the orbital distance of the planet’s Moon from the planet:
8.6
Once we have that we can get the radius of the Moon of the planet
8.7
Now we can get the rotation period of the planet, its length of day:
8.8
Because the orbital velocity of the planet’s Moon is
8.9
We can use the the delocalization time get the moon’s mass, which is 1/2 the planet’s year, with
8.10 , See Appendix 1
2
R
⊙
R
m
M
e
M
⊙
= 1
M
e
=
5
6
2
M
⊙
R
2
⊙
r
2
e
M
e
= 0.13889(1.9891E 30kg)
6.96E8m
2
1.496E11m
2
= 5.979748E 24kg
5.972E 24
5.979748E 24
100 = 99.87 %
r
planet
r
moon
≈
R
star
R
moon
2
R
⊙
R
m
M
e
M
⊙
= 1
Ear th Da y =
2r
e
v
m
M
m
M
⊙
3
v
m
=
GM
e
r
m
τ =
m
moon
(2r
moon
)
2
ℏ
s
of 44 56
And the whole system is solved. Let us see how the theoretical luminosity appears with that
measured with equation 8.1:
It is close to the measured. The equation for predicting is approximate as the actual
values can vary a little with things like metallicity of the stars. Tau Ceti has low metallicity
compared to our Sun.
Let’s find the orbital distance of the habitable planet if it exists with equation 8.2:
Let’s find the orbital period of the habitable planet (It’s year) with equation 8.3:
,
=
Let’s get the mass of the habitable planet from equation 8.5:
=
Which is good because the planet, if it exists, is thought to be larger than the Earth. Let’s get the
radius of its moon from equation 8.4:
=
Let’s see at what distance it orbits the planet with equation 8.6:
L
s
= M
3.5
s
= 0.783
3.5
= 0.425L
⊙
0.448L
⊙
r
p
= 0.488 = 0.69857AU = (0.6986AU )(1.496E11m /AU ) = 1.045E11m eters
T
2
p
= r
3
p
T
2
p
= (0.69867AU )
3
T
p
= 0.584years(365.25d a ys)((24hrs)(60min)(60sec) = 1849638secon d s
(0.584)(365.25d a ys) = 213.306d a ys
M
p
=
5
36
M
s
R
2
s
r
2
p
M
s
= (0.783)(1.9891E 30kg) = 1.5575E 30kg
R
s
= (0.793)(6.96E8m) = 5.52E8m
M
p
=
5
36
(1.5575E 30kg)
(5.52E8m)
2
(1.045E11m)
2
= 6.036E 24kg
6.036E 24
5.972E 24
= 1.0107Ear th Ma sses
R
m
= 2R
s
M
p
M
s
2(5.52E 8m)
(6.036E 24kg)
(1.5575E 30kg)
= 1.53656E6m =
1.53656E6m
1.7374E6m
= 0.8844Lun arRa dii
of 45 56
=
Let’s get the orbital velocity of this moon with equation 8.9:
So the orbital period of the moon is:
=
Compared to that of the Earth which is 27.3 days for the sidereal month and is a 29.53day lunar
month. The Moon of Tau Ceti would have a slightly longer lunar month, too. Now we can
determine the rotation period of the planet which would be its day:
We get the mass of the moon from equation 8.10:
=
So the rotation period of the planet (Its Day) is from equation 8.8:
=
=
If the moon of Tau Ceti is to perfectly eclipse its star, like our Moon does with the Sun, to let its
inhabitants know that they are there for reason, and so as to theoretically be a condition for
optimizing life, then we have the following should hold:
r
moon
= r
planet
R
moon
R
star
= (1.045E11m)
1.53656E6m
5.52E8m
= 2.91E8m
2.91E8m
3.84E8m
= 0.7578L u n ar Orbital Ra dii
v
m
=
GM
p
r
m
=
(6.67408E − 11)(6.0636E 24kg
2.91E8m
= 1,176.6m /s
T
m
=
2π r
m
v
m
=
2π (.91E 8m)
1,176.6m /s
= 1.554E6secon d s
(1.554E6s)
mi n
60sec
h our
60min
d a y
24h ours
= 17.987d a ys
M
m
=
ℏ
s
τ
4r
2
m
=
(2.8314E 33J ⋅ s)(0.5)(18429638s)
4(2.91E 8m)
2
= 7.7E 22kg
7.7E 22kg
7.34763E 22kg
= 1.048Ear th Moon s
Pl an et Day =
2r
p
v
m
M
m
M
s
3
2(1.045E11m)
1,176.6m /s
7.7E 22kg
1.5575E 30kg
= 39495.61secon d s = 658.26min = 10.971h ours
10.971
24h ours
= 0.457Ear th Days
r
planet
r
moon
≈
R
star
R
moon
of 46 56
We have
And it does hold. For the Earth/Moon/Sun system we get
For the Tau Ceti system the 359 is close to 360, which is a convenient amount of degrees into
which divide a circle, like we did. We see the Tau Cetians might do the same, because like us
they might choose the base 60 sexagesimal system of counting, and this 360 might influence the
way the inhabitants of a planet orbiting this star would make their calendar. Just like the 365
day year here on Earth corresponds closely to the 360 degrees of a circle. Aside from having here
on Earth the degree, we have gradians, of which there are 400 in a circle. They make it very easy
for computing right angles to a given angle. They are used more in Europe and in surveying and
architecture, though it might have some interesting connection with our 400=400.
The habitable planet of Tau Ceti would orbit it star with a period of 213 days, which would be its
year. It star is about 78% the mass of our Sun, and about 79% its size. Its habitable planet would
be a little larger than the Earth, but not much. Its Moon would be about 88% the size of our
moon, but a little bit more massive, but not much more massive. It would orbit Tau Ceti once
about every 18 days meaning there would be 11.833 months in the Tau Ceti Year, or about 12
months like we have. It would orbit the planet at about three quarters the distance ours does.
The day from sunrise to sunrise, or sunset to sunset would be about 11 hours, or close to half the
length of our day. This is if Tau Ceti was engineered for life, but it doesn’t have to have a planet
so optimally designed for life as ours. Our 24 hour day would be better for when entering the
realm of doing astronomy, because longer nights mean more complete observations of the
universe surrounding us than for the Tau Cetians. It may mean for the conditions for life to be
optimal, like here on Earth, the star would have to be as massive as the Sun, to provide a longer
year and a longer day. But we see here how the science of engineering planetary systems optimal
for life, might be done.
1.045E11m
2.91E8m
=
5.52E8m
1.53656E6m
359 = 359
400 = 400
of 47 56
9.0 A Genesis Project
Thus looking at the angular momentum equation for the Earth, we have
9.1
9.2
9.3
These yield:
9.4
Since
9.5
9.6
we can eliminate and have entirely classical values like mass, radius, KE, of Earth/Moon/Sun.
We have
9.7
Since we have:
9.8
This is the Earth orbit in terms of the angle theta between the Earth spin inclination and the
lunar orbital inclination. Let’s see what its accuracy is, using the data on page 36 of Part 3: Spin
we have:
=(1.2679E11m)(0.878)=1.077E11m
p
moon
r
earth
= K E
moon
(Ear th Da y) ⋅
M
⊙
M
m
⋅ n
−
1
4
cos(θ )
λ
moon
c
(K E
earth
) = K E
moon
(Ear th Da y)
K E
e
= n
R
⊙
R
m
⋅ G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
3
=
1
p
m
r
e
⋅
λ
m
c
M
⊙
M
m
⋅
R
2
⊙
R
2
m
⋅ GM
2
e
⋅ cos(θ )
λ
m
c
=
ℏ
⊙
GM
3
m
⋅
1
c
= 1secon d
K E
moon
K E
earth
(Ear th Da y) = 1secon d
ℏ
r
3
=
1
p
m
r
e
⋅
K E
moon
K E
earth
(Ear th Da y)
M
⊙
M
m
⋅
R
2
⊙
R
2
m
⋅ GM
2
e
⋅ cos(θ )
r
e
= r
3
= r
earth
r
2
3
=
1
p
m
⋅
K E
moon
K E
earth
(Ear th Da y)
M
⊙
M
m
⋅
R
2
⊙
R
2
m
⋅ GM
2
e
⋅ cos(θ )
r
3
=
1
1.1234E 37kg
m
2
s
⋅
3.837E 28J
2.7396E 33J
⋅ 86400s(5.03E 3) ⋅
(6.96E8m)
2
(1.7374E6m)
2
(6.67408E − 11)((5.97E 24kg)
2
cos(28.53
∘
)
of 48 56
Where we used the average orbital velocity of the Moon for the calculation of the momentum
1022 m/s. We use to get
=(2.06E11m)(0.878)=1.81043E11m
We average these two results to get
Since the average orbital distance of the Earth is , this is an accuracy of
Now we model in the standard model for radiative equilibrium of the Earth (As much radiation
entering equals as much leaving, that is the Earth is not cooling or warming) so we can
determine the annual average temperature of the Earth, its climate. We want to consider how
much sunlight reaches the Earth. The solar luminosity is:
9.9
The separation between the Earth and the Sun is
9.10
The solar luminosity is reduced at Earth by the inverse square law:
9.11
intercepts the Earth disc and distributes itself over the entire surface and because
the Earth’s albedo, , is 0.3 it reflects 30% of the light back into space. We have
9.12
Temperature to the fourth is proportional to radiation by the Steffan-Boltzmann constant, :
means temperature entering:
9.13
So we have the annual average temperature would be
v
ma x
= 1082m /s
r
3
=
1
1.8935E 37kg
m
2
s
⋅
3.837E 28J
.7396E 33J
⋅ 86400s(5.03E 3) ⋅
(6.96E8m)
2
(1.7374E6m)
2
(6.67408E − 11)((5.97E 24kg)
2
cos(28.53
∘
)
r
3
=
1.077E11m + 1.81043E11m
2
= 1.4437E11m
1AU = 1.496E11m
1.4437E11m
1.496E11m
100 = 96.5 %
L
0
= 3.9E 26J/s
r
e
= 1.5E11m
S
0
=
3.9E 26
4π (1.5E11m)
2
= 1370
Wat ts
m eter
2
S
0
π R
2
4π R
2
a
(1 − a)S
0
π R
2
4π R
2
= (1 − a)
S
0
4
σ
T
e
σ T
4
e
=
S
0
4
(1 − a)
of 49 56
9.14
The solar constant is usually 1361 (Well established) because we rounded our value for the Earth
distance and the Earth orbit is not perfectly circular. Let’s convert this to centigrade:
The actual annual average temperature of the Earth is 14C or 57F. But we have not included the
effect of the Earth’s greenhouse gases holding in heat. In a one layer atmosphere model we
consider that The radiation entering the system equals the radiation leaving the system when
the Earth is in radiative equilibrium. The atmosphere radiates back to the surface. So
the temperature at the surface of the Earth is
9.15
So,…
9.16
So this is hotter than the annual average temperature we said was 14C or 57F but we have not
considered cooling by convection; A lot of the radiation goes into warming the ocean, which has
a high specific heat (4.184J/g-K) and when this water evaporates it precipitates and returns as
rain. If we include this we get much closer to accurately predicting the global temperature. But
we are not going to consider things outside astronomy and are going to use this with our
equation 6 for the orbit of the Earth as determined by the Moon and Earth tilt, and suggest the
Moon plays and integral role in life on Earth, more so than we already realize.
We write
9.17
9.18
T
e
=
4
S
0
4σ
(1 − a)
σ = 5.67E − 8
T
e
=
4
1361
4(5.67E − 8)
(0.7) = 254.58K
T
e
= 254.58K − 273.15K = − 18.57
∘
C
(−18.57)(1.8) + 32 = − 1.42
∘
F
T
a
= T
e
σ T
4
S
= σ T
4
e
+ σ T
4
a
= 2σ T
4
e
T
S
= 2
1
4
T
e
T
S
= (1.18921)254.58 = 302.75K
T
S
= 302.75 − 273.15 = 29.6
∘
C
29.6(1.8) + 32 = 85.28
∘
F
S
0
=
L
o
4π r
2
3
T
e
=
4
L
0
4π r
2
3
4σ
(1 − a) =
1
4
L
0
π r
2
3
σ
(1 − a)
of 50 56
9.19
9.20
Where is the spin of the Earth, is one day, time for one revolution on its axis. We have
9.21
This is the annual average temperature of the Earth not compensated for non-astronomical
effects (convection, vaporization, and wind streams) rather is in terms of the angle between the
Earth tilt, and lunar orbit. It has a value of
Which is high, but there will be cooling by convection. This is actually cooler than the hottest
day on record on Earth, which was on 10 July 1913 at Greenland Ranch in
Death Valley, California, USA.
T
S
= 2
1
4
1
4
L
0
π r
2
3
σ
(1 − a)
r
2
3
=
1
p
m
r
e
⋅
K E
moon
K E
earth
(Ear th Da y)
M
⊙
M
m
⋅
R
2
⊙
R
2
m
⋅ GM
2
e
⋅ cos(θ )
r
e
= r
3
1
r
2
3
= p
m
K E
e
K E
m
⋅
1
S
e
M
M
M
⊙
⋅
R
m
R
⊙
⋅
1
GM
2
e
cos(θ )
S
e
T
S
=
2
1
4
4
⋅ p
m
K E
e
K E
m
⋅
1
S
e
M
M
M
⊙
⋅
R
m
R
⊙
⋅
1
GM
2
e
cos(θ )
⋅
L
0
πσ
(1 − a)
T
s
=
0.2973(1.1234E 37)
2.7396E 33J
3.837E 28J
⋅
1
86400s
7.347673E 22kg
1.9891E 30kg
(1.7374E6m)
2
(6.96E8m)
2
1
(6.67408E − 11)(5.972E 24kg)
2
(0.878)
×
3.9E 26J/s
π (5.67E − 8)
= (8.34025E − 15)(3.91485E16) = 326.51
∘
K = 53.36
∘
C = 128.048
∘
F
56.7
∘
C = 134
∘
F
of 51 56
10.0 Hydrocarbons The Backbones of Life Chemistry !
Interestingly we can write!
10.1
10.2
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting cancel with the bodies of these equations on the left, but rather
divide into them, which are in units of mass, giving us a number of protons. I say this is the
biological because these are the hydrocarbons the backbones of biological chemistry. We see
they display sixfold symmetry. I can generate the time values from these equations for all of the
elements for integer values in a program. Some of the output looks like
A
very
interesting thing here is looking at the values generated by the program, the smallest integer value 1
second produces 6 protons (carbon) and the largest integer value 6 seconds produces one proton
(hydrogen). Beyond six seconds you have fractional protons, and the rest of the elements heavier than
carbon are formed by fractional seconds. These are the hydrocarbons the backbones of biological
chemistry. And carbon is the core element of life. Here is the code for the program:
1
α
2
m
p
h 4π r
2
p
Gc
= 6pr oton ⋅ secon ds = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1pr oton ⋅ 6secon d s = hydr ogen(H )
m
p
of 52 56
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792458,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
of 53 56
Appendix 1
If we want to prove that our planetary Planck constant is correct, the delocalization time for the
Earth should be 6 months using it, the time for the Earth to travel the width of its orbit. We want
to solve the Schrödinger wave equation for a wave packet and use the most basic thing we can
which is a Gaussian distribution. We want to then substitute for Planck’s constant that is used
for quanta and atoms our Planck-type constant (h bar solar) for the Earth/Moon/Sun system
then apply it to predict the delocalization time for the Moon in its orbit with the Earth around
the Sun.
We consider a Gaussian wave-packet at t=0:
We say that is the delocalization length and decompose the wave packet with a Fourier
transform:
is the harmonics of the wave function. We use the identity that gives the integral of a
quadratic:
Solve the equation
With the initial condition
A plane wave is the solution:
Where,
The wave-packet evolves with time as
Calculate the Gaussian integral of
h
ℏ
⊙
ψ (x,0) = Ae
−
x
2
2d
2
d
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
ℏ
px
ϕ
p
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
iℏ∂
t
ψ (x, t) =
p
2m
ψ (x, t)
ψ (x,0) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
ℏ
px
e
i
ℏ
( px−ϵ( p)t )
ϵ( p) =
p
2
2m
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
ℏ
( px−
p
2
2m
t)
dp
of 54 56
and
The solution is:
where
is the delocalization distance, which for instance could be the width of an atom. is the
delocalization time, the average time for say an electron to traverse the diameter of the atom and
even leave it, to delocalize. If we substitute for our , and say that the delocalization distance
is for the Moon, the width of the Earth orbit, we should get a half a year for the delocalization
time, the time for the Moon and Earth to traverse the diameter of their orbit around the Sun. We
have
Where is the mass of the Moon, and is the orbital radius of the Moon. We have
Now let’s compute a half a year…
(1/2)(365.25)(24)(60)(60)=15778800 seconds
So we see our delocalization time is very close to the half year over which the Earth and Moon
travel from one position to the opposite side of the Sun. The closeness is
Thus we know our is accurate, it continues to function in a theoretical framework. The thing
about this is that it means we can predict the mass of the Moon from the Earth year. In terms of
what we said earlier that the Moon allows for life by creating the seasons, holding the Earth at
its tilt to the Sun, so we don’t go through extreme heat and cold, this suggests the Moon has a
mass that follows from Earth orbit, which is the habitable zone of the Sun, the right distance for
water to exist as liquid, and thus could be as it is for a reason, which means life might be part of
a physical process throughout the Universe, that it unfolds naturally in the evolution of star
systems.
α
2
=
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
ψ
2
= ex p
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
d
τ
ℏ
ℏ
⊙
τ =
m
moon
(2r
moon
)
2
ℏ
⊙
m
moon
r
moon
τ = 4
(7.34767E 22kg)(3.844E8m)
2
2.8314E 33J ⋅ s
= 15338227secon d s
15338227
15778800
100 = 97.2 %
ℏ
⊙
of 55 56
Some of the data used in this paper:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and
Earth’s orbital velocity at perihelion we have:
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G: 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,458m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 56 56
The Author
